3.735 \(\int \frac {e^{x^2} (1+4 x^4)}{x^2} \, dx\)

Optimal. Leaf size=19 \[ 2 e^{x^2} x-\frac {e^{x^2}}{x} \]

[Out]

-exp(x^2)/x+2*exp(x^2)*x

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Rubi [A]  time = 0.10, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6742, 2214, 2204, 2212} \[ 2 e^{x^2} x-\frac {e^{x^2}}{x} \]

Antiderivative was successfully verified.

[In]

Int[(E^x^2*(1 + 4*x^4))/x^2,x]

[Out]

-(E^x^2/x) + 2*E^x^2*x

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {e^{x^2} \left (1+4 x^4\right )}{x^2} \, dx &=\int \left (\frac {e^{x^2}}{x^2}+4 e^{x^2} x^2\right ) \, dx\\ &=4 \int e^{x^2} x^2 \, dx+\int \frac {e^{x^2}}{x^2} \, dx\\ &=-\frac {e^{x^2}}{x}+2 e^{x^2} x\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.84 \[ \frac {e^{x^2} \left (2 x^2-1\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*(1 + 4*x^4))/x^2,x]

[Out]

(E^x^2*(-1 + 2*x^2))/x

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fricas [A]  time = 0.39, size = 15, normalized size = 0.79 \[ \frac {{\left (2 \, x^{2} - 1\right )} e^{\left (x^{2}\right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*(4*x^4+1)/x^2,x, algorithm="fricas")

[Out]

(2*x^2 - 1)*e^(x^2)/x

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giac [A]  time = 0.21, size = 20, normalized size = 1.05 \[ \frac {2 \, x^{2} e^{\left (x^{2}\right )} - e^{\left (x^{2}\right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*(4*x^4+1)/x^2,x, algorithm="giac")

[Out]

(2*x^2*e^(x^2) - e^(x^2))/x

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maple [A]  time = 0.03, size = 16, normalized size = 0.84 \[ \frac {\left (2 x^{2}-1\right ) {\mathrm e}^{x^{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x^2)*(4*x^4+1)/x^2,x)

[Out]

exp(x^2)*(2*x^2-1)/x

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maxima [C]  time = 0.74, size = 36, normalized size = 1.89 \[ 2 \, x e^{\left (x^{2}\right )} + i \, \sqrt {\pi } \operatorname {erf}\left (i \, x\right ) - \frac {\sqrt {-x^{2}} \Gamma \left (-\frac {1}{2}, -x^{2}\right )}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x^2)*(4*x^4+1)/x^2,x, algorithm="maxima")

[Out]

2*x*e^(x^2) + I*sqrt(pi)*erf(I*x) - 1/2*sqrt(-x^2)*gamma(-1/2, -x^2)/x

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mupad [B]  time = 0.07, size = 15, normalized size = 0.79 \[ \frac {{\mathrm {e}}^{x^2}\,\left (2\,x^2-1\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2)*(4*x^4 + 1))/x^2,x)

[Out]

(exp(x^2)*(2*x^2 - 1))/x

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sympy [A]  time = 0.09, size = 12, normalized size = 0.63 \[ \frac {\left (2 x^{2} - 1\right ) e^{x^{2}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x**2)*(4*x**4+1)/x**2,x)

[Out]

(2*x**2 - 1)*exp(x**2)/x

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